What is the area of a triangle with sides of length 1, 2, and 2?

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What is the area of a triangle with sides of length 1, 2, and 2?

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Answer 1 · 2016-01-25T08:52:23

$A r e a = \frac{\sqrt{15}}{4}$ $Area=sqrt15/4$ square units

Explanation:

Use the Heron's formula

$A r e a = \sqrt{s \cdot (s - a) (s - b) (s - c)}$ $Area=sqrt(s*(s-a)(s-b)(s-c)$

Let $a = 2, b = 2, c = 1$ $a=2, b=2, c=1$

compute $s$ $s$

$s = \frac{1}{2} (a + b + c) = \frac{1}{2} \cdot (2 + 2 + 1) = \frac{5}{2}$ $s=1/2(a+b+c)=1/2*(2+2+1)=5/2$

Now, compute the Area

$A r e a = \sqrt{s \cdot (s - a) (s - b) (s - c)}$ $Area=sqrt(s*(s-a)(s-b)(s-c)$

$A r e a = \sqrt{\frac{5}{2} \cdot (\frac{5}{2} - 2) (\frac{5}{2} - 2) (\frac{5}{2} - 1)}$ $Area=sqrt(5/2*(5/2-2)(5/2-2)(5/2-1)$

$A r e a = \sqrt{\frac{5}{2} \cdot (\frac{1}{2}) (\frac{1}{2}) (\frac{3}{2})} = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4}$ $Area=sqrt(5/2*(1/2)(1/2)(3/2))=sqrt(15/16)=sqrt15/4$

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What is the area of a triangle with sides of length 1, 2, and 2?

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